Find the value of $c$ so that the polynomial $p(x)$ is divisible by $(x-3)$. $p(x) = -x^3+cx^2-4x+3$ $c=$
Explanation: The following statements are equivalent: $(x-3)$ is a factor of $p(x)$ $p(x)$ is divisible by $(x-3)$ The remainder of $\dfrac{p(x)}{x-3}$ is $0$ We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $p(x)$ is divided by $(x-{3})$ is equal to $p({3})$. We want this remainder to be equal to $0$. So let's set $p({3})=0$ and solve this equation to find $c$. Let's plug ${x=3}$ in $p( x) = - x^3+c x^2-4 x+3$ and set that equal to $0$. $\begin{aligned} -({3})^3+c({3})^2 -4( 3)+3&=0 \\\\ -27+9c-12+3&=0 \\\\ -36+9c&=0 \\\\ 9c&=36 \\\\ c&=4 \end{aligned}$ To conclude, $c=4$